Puzzle for May 20, 2023 ( )
Scratchpad
Find the 6-digit number ABCDEF by solving the following equations:
A, B, C, D, E, and F each represent a one-digit positive integer.
Scratchpad
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Hint #1
eq.5 may be written as: C + D + E = A + F - E In the above equation, replace A + F with B + D + E (from eq.2): C + D + E = B + D + E - E which becomes C + D + E = B + D Subtract D from each side of the above equation: C + D + E - D = B + D - D which becomes eq.5a) C + E = B
Hint #2
In eq.4, replace C + E with B (from eq.5a): A + B = B + F Subtract B from each side of the equation above: A + B - B = B + F - B which makes A = F
Hint #3
In eq.5, substitute A for F: C + D + E = A - E + A which becomes C + D + E = 2×A - E Add E to both sides of the above equation: C + D + E + E = 2×A - E + E which becomes eq.5b) C + D + 2×E = 2×A
Hint #4
Substitute A for F in eq.3: B + C = A + D + A which becomes eq.3a) B + C = 2×A + D
Hint #5
Substitute C + D + 2×E for 2×A (from eq.5b) into eq.3a: B + C = C + D + 2×E + D which becomes B + C = C + 2×D + 2×E Subtract C from both sides of the equation above: B + C - C = C + 2×D + 2×E - C which becomes B = 2×D + 2×E Divide both sides by 2: B ÷ 2 = (2×D + 2×E) ÷ 2 which makes eq.3b) ½×B = D + E
Hint #6
Substitute A for F, and ½×B for D + E (from eq.3b) into eq.2: A + A = B + ½×B which becomes 2×A = 1½×B Divide both sides of the above equation by 2: 2×A ÷ 2 = 1½×B ÷ 2 which makes A = ¾×B and also makes A = F = ¾×B
Hint #7
Substitute (¾×B) for A in eq.3a: B + C = 2×(¾×B) + D which becomes B + C = 1½×B + D Subtract B from each side of the equation above: B + C - B = 1½×B + D - B which becomes eq.3c) C = ½×B + D
Hint #8
Substitute ¾×B for A and F, and ½×B + D for C (from eq.3c) in eq.6: ¾×B × D = B + ½×B + D - D + ¾×B which becomes ¾×B × D = 2¼×B Divide both sides of the above equation by ¾×B: ¾×B × D ÷ ¾×B = 2¼×B ÷ ¾×B which makes D = 3
Hint #9
Substitute 3 for D in eq.3c: eq.3d) C = ½×B + 3
Hint #10
Substitute 3 for D in eq.3b: ½×B = 3 + E Subtract 3 from each side of the above equation: ½×B - 3 = 3 + E - 3 which becomes eq.3e) ½×B - 3 = E
Hint #11
Substitute ¾×B for A and F, ½×B + 3 for C (from eq.3d), 3 for D, and ½×B - 3 for E (from eq.3e) in eq.1: ¾×B + B + ½×B + 3 + 3 + ½×B - 3 + ¾×B = 31 which simplifies to 3½×B + 3 = 31 Subtract 3 from both sides of the above equation: 3½×B + 3 - 3 = 31 - 3 which makes 3½×B = 28 Divide both sides by 3½: 3½×B ÷ 3½ = 28 ÷ 3½ which means B = 8
Solution
Since B = 8, then: A = F = ¾×B = ¾×8 = 6 C = ½×B + 3 = ½×8 + 3 = 4 + 3 = 7 (from eq.3d) E = ½×B - 3 = ½×8 - 3 = 4 - 3 = 1 (from eq.3e) and ABCDEF = 687316